Question 667427
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \csc\left(\varphi\right)\ =\ \frac{1}{\sin\left(\varphi\right)}]


Hence, *[tex \LARGE \sin(x)\ =\ \frac{1}{4}]


But since *[tex \LARGE \cos\left(\varphi\right)^2\ +\sin\left(\varphi\right)^2\ =\ 1], *[tex \LARGE \cos(x)\ =\ \pm\frac{\sqrt{15}}{4}].


But since the tangent is the sine divided by the cosine and we have a positive sine value, and are given that the tangent is negative, the cosine must be negative. Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(x)\ =\ -\frac{\sqrt{15}}{4}]


Using the double angle formulae:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\left(2\varphi\right)\ =\ 2\sin\left(\varphi\right)\cos\left(\varphi\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\left(2x\right)\ =\ 2\left(\frac{1}{4}\right)\left(-\frac{\sqrt{15}}{4}\right)\ =\ -\frac{\sqrt{15}}{8}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\left(2\varphi\right)\ =\ \cos^2\left(\varphi\right)\ -\ \sin^2\left(\varphi\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\left(2x\right)\ =\ \frac{15}{16}\ -\ \frac{1}{16}\ =\ \frac{7}{8}]


Then since:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan\left(\varphi\right)\ =\ \frac{\sin\left(\varphi\right)}{\cos\left(\varphi\right)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan\left(2x\right)\ =\ \frac{\sin\left(2x\right)}{\cos\left(2x\right)}\ =\ \frac{\frac{-\sqrt{15}}{8}}{\frac{7}{8}}\ =\ -\frac{\sqrt{15}}{7}]

 


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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